Trong LOGO để viết lên màn hình dòng chữ “Xin chào Việt Nam”, ta dùng câu lệnh:

A. LABEL { Xin chào Việt Nam}
C. PRINT {Xin chào Việt Nam}
B. LABEL [Xin chào Việt Nam]
D. PRINT [Xin chào Việt Nam]

đáp án A.

D

D

D 

B

chào bạn

chào

a.\(A=\dfrac{1}{x-1}-\dfrac{x^2+x}{x^2+1}.\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)\);\(ĐK:x\ne\pm1\)

\(A=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\left(\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\right)\)

\(A=\dfrac{1}{\left(x-1\right)}-\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)

\(A=\dfrac{1}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x^2+1\right)}\)

\(A=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\)

\(A=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}\)

\(A=\dfrac{x-1}{x^2+1}\)

b.\(A=0,2=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{x-1}{x^2+1}=\dfrac{1}{5}\)

\(\Leftrightarrow x^2+1=5x-5\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

c.\(A< 0\) mà \(x^2+1\ge1>0\)

--> A<0 khi \(x-1< 0\)

                  \(\Leftrightarrow x< 1\)

a. -ĐKXĐ:\(x\ne\pm1\)

\(A=\dfrac{1}{x-1}-\dfrac{x^2+x}{x^2+1}.\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\left(\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}\)

\(=\dfrac{x^2+1}{\left(x^2+1\right)\left(x-1\right)}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\)

b. \(A=\dfrac{x-1}{x^2+1}=0,2\)

\(\Leftrightarrow\dfrac{x-1}{x^2+1}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{5\left(x-1\right)}{5\left(x^2+1\right)}=\dfrac{x^2+1}{5\left(x^2+1\right)}\)

\(\Rightarrow5x-5=x^2+1\)

\(\Leftrightarrow x^2-5x+1+5=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

c. \(A=\dfrac{x-1}{x^2+1}< 0\)

\(\Leftrightarrow x-1< 0\) (vì \(x^2+1>0\forall x\))

\(\Leftrightarrow x< 1\)

Câu 1:

a: \(\frac37+\left(-\frac52\right)+\frac35\)

\(=\frac37-\frac52+\frac35\)

\(=\frac{30}{70}-\frac{175}{70}+\frac{42}{70}=\frac{30+42-175}{70}=-\frac{103}{70}\)

b: \(0,8-\left(-\frac27\right)-\frac{7}{10}=\frac45+\frac27-\frac{7}{10}\)

\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}=\frac{56+20-49}{70}=\frac{56-29}{70}=\frac{27}{70}\)

c: \(\frac34-\left\lbrack\left(-\frac53\right)-\left(\frac{1}{12}+\frac29\right)\right\rbrack\)

\(=\frac34+\frac53+\frac{1}{12}+\frac29\)

\(=\frac{27}{36}+\frac{60}{36}+\frac{3}{36}+\frac{8}{36}=\frac{98}{36}=\frac{49}{18}\)

d: \(19\frac13\cdot\frac45-\frac45\cdot15\frac13\)

\(=\frac45\left(19+\frac13-15-\frac13\right)\)

\(=\frac45\cdot4=\frac{16}{5}\)

e: \(\frac38\cdot\left(-\frac{3}{11}\right)+\frac38\cdot\left(-\frac{8}{11}\right)+2\frac57\)

\(=\frac38\left(-\frac{3}{11}-\frac{8}{11}\right)+2+\frac57\)

\(=-\frac38+2+\frac57=\frac{13}{8}+\frac57=\frac{91}{56}+\frac{40}{56}=\frac{131}{56}\)

f: \(\left(-\frac{5}{11}:\frac{13}{8}-\frac{5}{11}:\frac{13}{3}\right)-\left|-\frac{1}{33}\right|\)

\(=-\frac{5}{11}\cdot\frac{8}{13}-\frac{5}{11}\cdot\frac{3}{13}-\frac{1}{33}\)

\(=-\frac{5}{11}\cdot\frac{11}{13}-\frac{1}{33}=-\frac{5}{13}-\frac{1}{33}=\frac{-165-13}{13\cdot33}=\frac{-178}{429}\)

g: \(\frac{4}{23}+\frac{5}{21}+\frac12-\frac{4}{23}+\frac{16}{21}\)

\(=\left(\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac12\)

\(=1+\frac12=\frac32\)

h: \(\frac{13}{25}+\frac{6}{41}+\frac{-38}{25}+\frac{35}{41}-0,5\)

\(=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-0,5\)

\(=-\frac{25}{25}+\frac{41}{41}-0,5=-0,5\)

i: \(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{15}{17}+\frac23\)

\(=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac13+\frac23\right)-1-\frac{15}{17}\)

\(=1+1-1-\frac{15}{17}=1-\frac{15}{17}=\frac{2}{17}\)

j: \(\left(\frac{2022}{2023}\right)^0-\left(-3\right)_{}^3+\frac12:\left(-\frac12\right)^3\)

\(=1-\left(-27\right)+\frac12:\frac{-1}{8}\)

=28-4

=24

k: \(\left(-2\right)^3+\frac12:\frac{-1}{4}-\left|15\right|\)

\(=-8+\frac12\cdot\left(-4\right)-15\)

=-8-2-15

=-25

l: \(3:\left(-\frac32\right)^2+\frac19\cdot\sqrt{36}\)

\(=3:\frac94+\frac19\cdot6=3\cdot\frac49+\frac69=\frac{12}{9}+\frac69=\frac{18}{9}\)

=2

m: \(\sqrt{81}\cdot\frac13-\left(-3\right)^3=9\cdot\frac13-\left(-27\right)\)

=3+27

=30

n: \(\sqrt{\frac{16}{25}}\cdot\sqrt{\frac{121}{64}}-1\frac{3}{10}\)

\(=\frac45\cdot\frac{11}{8}-\frac{13}{10}\)

\(=\frac{11}{10}-\frac{13}{10}=-\frac{2}{10}=-\frac15\)

o: \(\frac{3^2}{2^3}\cdot\sqrt{144}-50\%\)

\(=\frac98\cdot12-\frac12=9\cdot\frac32-\frac12=\frac{27}{2}-\frac12=\frac{26}{2}\)

=13

p: \(\frac{6^2+3\cdot6^2+3^2}{-13}\)

\(=\frac{36+3\cdot108+9}{-13}=\frac{45+324}{-13}=\frac{369}{-13}=-\frac{369}{13}\)

q: \(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{\left(5\cdot20\right)^4}{\left(25\cdot4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)

r: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot5}=\frac23\cdot\frac65=\frac{12}{15}=\frac45\)

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

IX
2. j
3. i
4. f
5. c
6. a
7. h
8. e
9. g
10. d

XI
2. part => parts
3. a => an
4. a => an
5. a => the
6. are => will be (không chắc lắm)
7. taking => take
8. are => is

C. 
Bài 1
1. C
2. B
3. C
4. B

(Nên double-check trước khi chép)
 

Mik xin cảm ơn bn nha!!!

chào bạn